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slamgirl [31]
3 years ago
7

A 250-ml sample of na2so4 is reacted with an excess of bacl2. if 5.28 g baso4 is precipitated, what is the molarity of the na2so

4 solution?
Chemistry
1 answer:
Oksanka [162]3 years ago
3 0
To get the molarity you need to follow this equation
                       moles of solute
Molarity (M = -----------------------
                        Liters of solution

But before you apply that equation you need to find the moles of solute and the liters of solution. Follow this equation

Na2SO4 + BaCl2 = BaSO4 + 2 NaCl

Solution

Moles of BaSO4 = 5.28 g 
                               ---------------
                                233.43 g / mol
                             =  0.0226  moles
Moles of NaSO4 = 0.0226 
                  0.0226 mole
Molarity = -----------------
                  0.250 L
              =  0.0905 mol / L

So the answer is 0.0905 mol / L
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4 0
4 years ago
Answer asap with at least 3 or more sentences!
mixas84 [53]

Answer:

no.

Explanation:

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6 0
3 years ago
Read 2 more answers
An increase in temperature affects the reaction rate by &gt;decreasing the velocities of particles that collide in the reaction.
Bond [772]

Answer:

increasing the number of molecules that have sufficient kinetic energy to react.

Explanation:

An increase in temperature affects the reaction rate by increasing the number of molecules that have sufficient kinetic energy to react.

or we say; temperature increase, leads to an increase in the amount of collisions between molecules.

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A tank contains 90 kg of salt and 2000 L of water. Pure water enters a tank at the rate 6 L/min. The solution is mixed and drain
sashaice [31]

Answer:

a) 90 kg

b) 68.4 kg

c) 0 kg/L

Explanation:

Mass balance:

-w=\frac{dm}{dt}

w is the mass flow

m is the mass of salt

-v*C=\frac{dm}{dt}

v is the volume flow

C is the concentration

C=\frac{m}{V+(6-3)*L/min*t}

-v*\frac{m}{V+(6-3)*L/min*t}=\frac{dm}{dt}

-3*L/min*\frac{m}{2000L+(3)*L/min*t}=\frac{dm}{dt}

-3*L/min*\frac{dt}{2000L+(3)*L/min*t}=\frac{dm}{m}

-3*L/min*\int_{0}^{t}\frac{dt}{2000L+(3)*L/min*t}=\int_{90kg}^{m}\frac{dm}{m}

-[ln(2000L+3*L/min*t)-ln(2000L)]=ln(m)-ln(90kg)

-ln[(2000L+3*L/min*t)/2000L]=ln(m/90kg)

m=90kg*[2000L/(2000L+3*L/min*t)]

a) Initially: t=0

m=90kg*[2000L/(2000L+3*L/min*0)]=90kg

b) t=210 min (3.5 hr)

m=90kg*[2000L/(2000L+3*L/min*210min)]=68.4kg

c) If time trends to infinity the  division trends to 0 and, therefore, m trends to 0. So, the concentration at infinit time is 0 kg/L.

6 0
3 years ago
Is this molecule polar or nonpolar?
lutik1710 [3]
The answer is nonpolar  

3 0
3 years ago
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